Worked solution · Exam P

SOA Exam P Sample Question #612 Solution

This Exam P sample reference tests Exponential Distribution. The median identifies the exponential rate as ln(2)/10. Weighting the two disjoint failure-time windows by their respective payments gives 13.06384325, which rounds to choice A.

Independent analysis by ActuaryProof · reviewed July 30, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Exponential Distribution question

Setup

Use the median condition to determine the exponential rate.

\Pr(T>10)=e^{-10c}=0.5

c=\frac{\ln2}{10}

Model

Separate the two payment levels into disjoint time intervals.

E[W]=35\Pr(T\le5)+25\Pr(5<T\le7.5)

F(t)=1-e^{-ct}

Compute

Evaluate the exponential probabilities at both warranty boundaries.

E[W]=35F(5)+25\bigl(F(7.5)-F(5)\bigr)

E[W]=35(1-2^{-1/2})+25(2^{-1/2}-2^{-3/4})

E[W]=13.0638432506\ldots

Answer

The expected warranty payment rounds to 13.06.

\boxed{13.06\quad\text{(A)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

BThis overlaps the two payment windows by treating failures through 7.5 years as eligible for the first payment while also adding the later-window payment.
CThis adds a payment-weighted survival probability rather than using the probability of failure inside each covered interval.
DThis is close to 35P(T>5), which reverses failure and survival at the first warranty boundary and ignores the second payment tier.
EThis extends the second payment beyond the end of the warranty instead of assigning zero after 7.5 years.

Original practice · fully worked

Original variant: capped operating credits

A field instrument remains operational for an exponential time T with mean 4 years. A research program earns 12 credits for each year of operation, but only during the instrument's first three years. The credited amount is proportional for partial years. Calculate the expected total credit.

A
5.668
B
6.332
C
25.326
D
36.000
E
48.000

Answer and calculation

The credited operating time is min(T,3). Integrating its survival function gives 4(1-exp(−0.75)) years, so the expected credit is 48(1-exp(−0.75))=25.32640547 and choice C is correct.

Setup

Represent the credited duration by the operating lifetime capped at three years.

T\sim\operatorname{Exponential}\!\left(\text{mean }4\right)

C=12\min(T,3)

Model

Use the survival-integral identity for a capped nonnegative duration.

E[\min(T,3)]=\int_0^3\Pr(T>t)\,dt

\Pr(T>t)=e^{-t/4}

Compute

Evaluate the capped lifetime integral and multiply by the annual credit rate.

E[C]=12\int_0^3e^{-t/4}\,dt

E[C]=48(1-e^{-0.75})=25.3264054684\ldots

Answer

The expected total credit rounds to 25.326.

\boxed{25.326\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: capped operating credits

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables